Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{-5q - 35}{5q^2 + 50q + 45} \times \dfrac{q^2 - 3q - 4}{q - 4} $
First factor out any common factors. $a = \dfrac{-5(q + 7)}{5(q^2 + 10q + 9)} \times \dfrac{q^2 - 3q - 4}{q - 4} $ Then factor the quadratic expressions. $a = \dfrac {-5(q + 7)} {5(q + 1)(q + 9)} \times \dfrac {(q + 1)(q - 4)} {q - 4} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {-5(q + 7) \times (q + 1)(q - 4) } { 5(q + 1)(q + 9) \times (q - 4)} $ $a = \dfrac {-5(q + 1)(q - 4)(q + 7)} {5(q + 1)(q + 9)(q - 4)} $ Notice that $(q + 1)$ and $(q - 4)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-5\cancel{(q + 1)}(q - 4)(q + 7)} {5\cancel{(q + 1)}(q + 9)(q - 4)} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $a = \dfrac {-5\cancel{(q + 1)}\cancel{(q - 4)}(q + 7)} {5\cancel{(q + 1)}(q + 9)\cancel{(q - 4)}} $ We are dividing by $q - 4$ , so $q - 4 \neq 0$ Therefore, $q \neq 4$ $a = \dfrac {-5(q + 7)} {5(q + 9)} $ $ a = \dfrac{-(q + 7)}{q + 9}; q \neq -1; q \neq 4 $